Not even sure how to start on this one, walking me through completing the square would be helpful. I'm sure after that I can just apply a trig sub and finish the problem. It'd be extra cool if you walked me through the problem, though.
The problem:
Evaluate
$$ \int\frac{x^2}{\sqrt{4x-x^2}} \text{d}x $$
by first completing the square.
$$\int\frac{x^2}{\sqrt{4x-x^2}} dx$$ $$=\int\frac{x^2}{\sqrt{4-(x-2)^2}} dx$$ We now apply the substitution $x-2=u$ $$=\int\frac{(u+2)^2}{\sqrt{4-u^2}} du$$ The denominator is now ready for a simple trig substitution $u= 2\sin(t)$
If $-(a+x)^2=C+4x-x^2$ what would the value of $a$ need to be ? \begin{eqnarray*} 4x-x^2=4-(4-4x+x^2)=4-(2-x)^2=2^2-(2-x)^2 \end{eqnarray*} How does help with you're integral ? \begin{eqnarray*} \int \frac{x^2 \ dx}{ \sqrt{4x-x^2}}=\int \frac{x^2 \ dx}{\sqrt{ 2^2-(2-x)^2}} \end{eqnarray*} We would like the content of the square root to look like $1-\cos^2 \theta$. How will we do this ?
Let $2-x=2 \cos \theta \ $ so $ dx= \sin \theta \ d\theta \ $ & our integral becomes \begin{eqnarray*} \int \frac{x^2 \ dx}{ \sqrt{4x-x^2}}=\frac{4}{2} \int (1- \cos \theta)^2 \ d\theta \end{eqnarray*} Now expand the bracket & recall that the double angle formula for $\cos \theta$ gives $\cos^2 \theta=\frac{1+ \cos 2\theta}{2}$ now integrate term by term \begin{eqnarray*} 2 \int (1- 2\cos \theta +\cos^2\theta ) d\theta =2 \left(\theta - 2\sin \theta + \frac{1}{2}\theta + \frac{1}{4} \sin 2\theta \right)+c \end{eqnarray*} Now substitute back in to replace $\theta$ by $x$ and we have \begin{eqnarray*} \int \frac{x^2 \ dx}{ \sqrt{4x-x^2}}=3 \cos^{-1}( \frac{2-x}{2}) -2\sqrt{4x-x^2} + \frac{(2-x)}{4} \sqrt{4x-x^2} +c \end{eqnarray*}